If `f(x)=x^(2)+kx+1`, for all x and if it is an even function, find k.

Answer

To solve the problem, we need to find the value of \( k \) such that the function \( f(x) = x^2 + kx + 1 \) is an even function.

### Step-by-step Solution:

1. **Understand the definition of an even function**:
A function \( f(x) \) is considered even if it satisfies the condition \( f(x) = f(-x) \) for all \( x \).

2. **Write down the function**:
We have the function \( f(x) = x^2 + kx + 1 \).

3. **Calculate \( f(-x) \)**:
Substitute \(-x\) into the function:
\[ f(-x) = (-x)^2 + k(-x) + 1 \]
Simplifying this gives:
\[ f(-x) = x^2 - kx + 1 \]

4. **Set up the equation for evenness**:
For \( f(x) \) to be even, we need:
\[ f(x) = f(-x) \]
This means:
\[ x^2 + kx + 1 = x^2 - kx + 1 \]

5. **Eliminate common terms**:
We can subtract \( x^2 + 1 \) from both sides:
\[ kx = -kx \]

6. **Combine like terms**:
Rearranging gives us:
\[ kx + kx = 0 \implies 2kx = 0 \]

7. **Solve for \( k \)**:
Since this equation must hold for all \( x \), we can conclude that:
\[ k = 0 \]

### Final Answer:
Thus, the value of \( k \) is \( 0 \).

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